Integration can be used to find area under a curve. The value of the integral is an area if the integrand is always nonnegative on an interval.

$ \text{If}f(x) \geq 0 \text{ then } \text{ on }(a,b) \int_a^b f(x) \, dx \text{ $\geq $ 0.} $

</center> $ \text{Find the area of the region bounded by} y = x^3 \text{ and the lines } x = 2, x = 3. $

In [7]:
Plot[x^3, {x, 2, 3}, Filling-> Axis]
Out[7]:
Output
In [1]:
Integrate[x^3, {x, 2, 3}]
Out[1]:
Output

Integrals can also have a value which is negative or zero.

$ \text{Two examples of integrals which evaluate to zero.} $

$ \int_2^2 x^3 \, dx, $

$ \int _0^{\pi } \cos(x)\,dx. $

In [9]:
Plot[Cos[x], {x, 0, Pi}, Filling-> Axis]
Out[9]:
Output
In [2]:
Integrate[x^3, {x, 2, 2}]
Out[2]:
0

Integrals can be evaluated by using the geometric interpretation of the integral to simplify the calculations in some cases.

$ \text{For example, to evaluate} $

$ \int _{-2}^3|x|dx, $

$ \text{one would look at the graph of }|$x$|. $

In [3]:
Plot[{Abs[x], y = 0},{x, -2, 3}, PlotRange->Automatic, Filling-> Axis]
Out[3]:
Output

$ \text{Now add the areas of the two isosceles right triangles to evaluate the integral.} $

$ \int _{-2}^3|x|dx \text{= } \frac{1}{2} 2 \text{(2) + } \frac{1}{2} \text{(3)(3) =} \frac{13}{2} . $

$ \text{The integral can be used to find the average value of a function over an integral.} $

$ \text{Find the average value of f(x) = } e^x \text{ on the interval }[1, 3]. $

In [10]:
Plot[E^x, {x, 1, 3}, Filling-> Axis]
Out[10]:
Output
In [4]:
Integrate[E^x, {x, 1, 3}]
Out[4]:
Output

Symmetry can be used to simplify integrals.

$ \text{If }f\text{ is an } \text{even} \text{ function } \text{ then } $

$ \int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx. $
$ \text{If } f \text{ is } \text{ odd } \text{ then } $

$ \int_{-a}^a f(x) \, dx = 0. $
$ \text{Find }\int _{-1}^1x^2+x^3 dx. $

$ \text{First } \text{ note } \text{ that } x^2 \text{ an } \text{a nd } \text{ even } \text{ function } \text{ is } x^3 \text{ is an odd function, so that}\\ {\int _{-1}^1x^2+x^3 dx \text{= 2} \int_0^1 x^2 \, dx \text{= 2 (1/3) = 2/3.}} $

$\text{Check by using}$ Mathematica.

In [11]:
Plot[x^2 + x^3, {x, -1, 1}, Filling-> Axis]
Out[11]:
Output
In [12]:
Plot[x^2 , {x, -1, 1}, Filling-> Axis]
Out[12]:
Output
In [13]:
Plot[x^3, {x, -1, 1}, Filling-> Axis]
Out[13]:
Output
In [1]:
Integrate[x^2 + x^3, {x, -1, 1}]
Out[1]:
Output

An integral can be used to find the value of a function at a point.

$ \text{If the derivative is given as f'(x) and f(a) is known, then the Fundamental Theorem of Calculus says that } $

$ f(b) - f(a) = \int_a^b x f' \, dx \text{, so that f(b) = f(a) +} \int _a^bf'xdx. $

$ \text{Find the approximate value of }y\text{ at }x = 3\text{ , if } \frac{\text{d}y}{\text{d}x} = \text{Sin}(x)^3 \text{ and } y = 2 \text{ when } x = 1. $

In [2]:
2 + NIntegrate[(Sin[x])^3, {x,1,3}, WorkingPrecision->20]
Out[2]:
3.1542929547444195867

An Integral can be used to define a new function.

$ \text{One of the most notable examples is the new function, }$\text{ln}(x)$\text{ for logarithms with base }e. \text{The definition of ln is} $

$ \text{ln}(x) = \int _1^x\frac{1}{t}dt. $
Then
$ \text{ln}(2) = \int_1^2 \frac{1}{t} \, dt $
and
$ \frac{d}{\text{dx}}\text{ln}(x) = \frac{\text{d}}{\text{d}x} \int_1^x \frac{1}{t} \, \text{d}t = \frac{1}{x}. $

$ \text{Let g}(x) = \int _0^xt^3 \cos \text{d}t. $

$ \text{What is the maximum value of } g \text{on the interval [ 0, 2 ] ?} $

$\text{Note g}(0) = 0\text{ and }\text{g'}(x) = \cos(x^3).\text{ Since }\cos(x)\text{ is zero at }\frac{\pi }{2} \text{, to find the zero of g'}(x)\text{ take the cube root of }\frac{\pi }{2}.$

In [3]:
N[(Pi/2)^(1/3)]
Out[3]:
1.16245

The integral is an infinite sum of products. (Riemann Sum)

$ \text{Find} \lim_{n\to \infty } \frac{n}{n^2} + \frac{n}{(n+1)^2} +\cdots + \frac{n}{(2 n)^2}. $

$\text{Note that the sum can be rewritten as a product by multiplying by } \frac{n}{n}. \text{This gives the equivalent problem:} \lim_{n\to \infty }[\left(\frac{n}{n+1}\right)^2 +\left(\frac{n}{n+2}\right)^2 + \cdots + \left(\frac{n}{2 n}\right)^2] \frac{1}{n} = \lim_{n\to \infty } \sum _{k=1}^n \left(\frac{1}{\frac{k}{n}+1}\right)^2 \frac{1}{n} = \int_1^2 \frac{1}{x^2} \, \text{d}x.$

In [4]:
Integrate[1/x^2, {x, 1, 2}]
Out[4]:
Output

Exercises

$ \text{1. Find the area of the region bounded by }x = 1, x = 2, f(x) = \frac{1}{x^3} \text{ and the }x\text{-axis. } $

$ \text{2. Evaluate the following: a) } \int_1^3 (x-2) \, \text{d}x, \text{ b) } \int_3^0 (x-2) \, \text{d}x \text{, c) } \int_0^{2 \pi } x \cos \, \text{d}x \text{, d) }\int_{\pi }^0 x \sin \, dx. $

$ \text{3. Use the} \text{geometric interpretation of the integral to evaluate } \int _{-1}^3|x-2|\text{d}x. $

$ \text{4. Use the geometric interpretation of the integral to evaluate} \int _0^2\sqrt{4-x^2}\text{d}x. $

$ \text{5. What is the average value of f}(x) = x + \sin x \text{ on the interval }[ 0, \pi ]\text{?} $

$ \text{6. Use symmetry to evaluate }\int_{-\pi }^{\pi } \left(x^3+x+x \sin -x \cos +3\right) \, \text{d}x. $

$ \text{7. If the velocity of an object is v}{(t) = -16t + 56}\\ \text{ and the object is 5 feet off the ground at }t\text{ = 0, then how high will the object be seven seconds from now?} $

$ \text{8. Wine is leaking from a vat at the rate of } \text{R}{(t) = 1000 e^{-0.1 t}} \text{ gallons per hour, } \text{where }t\text{ is measured in hours. }\\ \text{How much wine has leaked out of the vat after 12 hours?} $

$ \text{9. If f}(0) = 1 \text{ and f'}(x) = \tan(x^3 \text{), then what is f}(1)\text{?} $

$ \text{10. If g}{(x) = \int _0^xt^2 \sin \text{d}t}\\ \text{ on the closed interval }{[ 0, 3 ]}\text{, then for what value of }x\text{ does g and for what }x\text{ does }g\text{ have a local maximum?}\\ \text{Plot the graph to verify your results.} $

$ \text{11. Find } {\lim_{n\to \infty }[\frac{n^2}{(n+1)^3}+\frac{n^2}{(n+2)^3}+\frac{n^2}{(2 n)^3}+\cdots].} $

$\text{Hint: Divide by}n^3 \text{and factor out the}\frac{1}{n} \text{ to form a Riemann sum.}$